Explanation Given Information: The unstretched length of spring is 0.3 m. The mass of each bar is 20 kg. Assumptions: Consider the state of member as tension where the force is pulling the member and as compression where the force is pushing the member. Consider the force indicating right side as positive and left side as negative in horizontal components of forces. Consider the force indicating upside is taken as positive and downside as negative in vertical components of forces. Consider clockwise moment as negative and anti-clock wise moment as positive wherever applicable. Apply the Equation of equilibrium wherever applicable. The acceleration due to gravity ( g ) is 9.81 m/sec 2 . Convert the mass of bar in to weight of bar using the relation. W = m g Here, the mass is m . The weight of each bar is ( 20 × 9.81 ) N . Member BC : Show the free body diagram of the member BC as in Figure (1). Using Figure (1), Spring force: Determine the elongation of the spring, x using the equation. x = 2 d − Unstretched length (I) Determine the spring force using the equation. F s p = k x (II) Here, the stiffness is k and the compression is x . Moment about point C : Determine the horizontal and vertical reaction at point B by taking moment about point C. ∑ M C = 0 B y ( 2 cos θ ) − B x ( 2 sin θ ) − 20 × 9.81 ( cos θ ) = 0 2 B y cos θ − 2 B x sin θ = 196.2 cos θ (III) Conclusion: Substitute 2 sin θ for d and 0.3 m for unstretched length in Equation (I). x = 2 ( 2 sin θ ) − 0 .3 = 4 sin θ − 0.3 Substitute 150 N/m for k and ( 4 sin θ − 0.3 ) for x in Equation (II). F s p = 150 × ( 4 sin θ − 0

INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)

14th Edition

ISBN: 9780133918922

Author: Russell C. Hibbeler

Publisher: PEARSON

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Free Body Diagram

The system of forces acting on a body tends to either rotate it or make the body undergo motion. In general, when an external agent exerts a force on a body, the body undergoes translational motion and rotational motion. The body has six degrees of freed…

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Chapter 6.6, Problem 110P

To determine

The angle θ for equilibrium.

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Chapter 6.6, Problem 109P

Chapter 6.6, Problem 111P

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INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)

Ch. 6.3 - In each case, calculate the support reactions and...Ch. 6.3 - Identify the zero-force members in each truss....Ch. 6.3 - State if the members are in tension or...Ch. 6.3 - State if the members are in tension or...Ch. 6.3 - State if the members are in tension or...Ch. 6.3 - Determine the greatest load P that can be applied...Ch. 6.3 - Identify the zero-force members in the truss....Ch. 6.3 - State if the members are in tension or...Ch. 6.3 - Set P1 = 20 kN, P2 = 10 kN. Probs. 6-1/2 Ch. 6.3 - Set P1 = 45 kN, P2 = 30 kN. Probs. 6-1/2

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The angle θ for equilibrium.

Solution Summary: The author explains the angle theta for equilibrium. The unstretched length of spring is 0.3 m and the mass of each bar is 20 kg.

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The angle θ for equilibrium.

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